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From my understanding, most forces that are conservative are of the form

$$\vec F = \hat i F(x)$$

Which means the force is only a function of one variable, which means the work done of the force in any direction $dr$ is independent of the path taken, as the integral is a function of $x$ only. For example, let $F(x) = x$, and $\vec {dr} = dx \hat i + dy \hat j$.

$$\therefore W = \int_{r_1}^{r_2}\vec F \cdot \vec dr = \int_{x_1}^{x_2}x\ dx \ + 0$$

$$W = \left[x^2/2\right]_{x_1}^{x_2}$$

So the work done only matters on the distance taken in only the direction along the force. However, is this force conservative?

$$\vec F = \hat i F(x) + \hat j F(y)$$

Let $F(x) = x$ and $F(y) = y$, then the work done is:

$$W = \int_{x_1}^{x_2} x \ dx + \int_{y_1}^{y_2} y \ dy $$

$$W = \left[x^2/2\right]_{x_1}^{x_2} + \left[y^2/2\right]_{y_1}^{y_2}$$

Here I think this is in indeed a conservative force because there is no integral with a function of a variable that isn't directly able to be integrated.

But, I think my lack of sureness with the previous statement above can be highlighted with the following example, where the next force I consider is not a conservative force.

$$F = \hat j F(x,y) = \hat j xy^2$$

$$W = \int_{y_1}^{y_2} xy^2\ dy$$

Here, the work done depends on how $x$ depends on $y$, which means the path taken does indeed matter.

(Note: I originally was going to want to express the work vector as a vector with component $\hat i$ instead of $\hat j$ to have an expression for work as $\int_{x_1}^{x_2} xy^2 \ dx$ and then $y$ would have to be expressed as a function of $x$ which sounds more intuitive to me but I wanted to do it this way to test my general understanding).

So am I right in implying that the force is only not conservative if any of the work integrals have a integral that is not expressed in terms of the differential of the variable, i.e integrating something like $xy^2$ in terms of $x$?

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  • $\begingroup$ Not an answer, just a comment on coordinate systems. If you wrote your co-ords in spherical, with conservative force, would that resolve things? Based on the fact that there is a central force involved somewhere. $\endgroup$ – user171879 Oct 28 '17 at 14:51
  • $\begingroup$ I was thinking it might resolve whether if the force had two vector components it could still be conservative, as that is the same as saying $\vec {F_r} = \hat r F(r)$ $\endgroup$ – sangstar Oct 28 '17 at 14:52
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The following are equivalent if the space is simply connected (no holes, one piece) and $U$ is not a function of time:

1. The force $\vec F$ is conservative.

2. $\nabla\times \vec F=0$

3. $W=\oint_\mathcal{C}\vec F\cdot\mathrm{d}\vec r=0\quad \forall\> \mathcal{C}$

4. $\vec F=-\nabla U$

Hope that it helps. (Might want to consider using another way than calculating integrals.)

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  • $\begingroup$ That is true if the space is simply connected and if $U$ does not depend explicitly on time. $\endgroup$ – Diracology Oct 28 '17 at 15:34
  • $\begingroup$ Of course, but it is certainly the case in these simple examples. I edited my answer. $\endgroup$ – Soap312 Oct 28 '17 at 15:38
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    $\begingroup$ The integral version (#3) is equivalent to others only if it vanishes over any arbitrary path/ contour. Since proving for an arbitrary path might prove tricky it is usually used to show that a force isn’t conservative by providing a contradicting example. $\endgroup$ – Yair M Oct 28 '17 at 15:56

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